//
// Created by Administrator on 2021/5/5.
//

/*
给你一个链表，删除链表的倒数第n个结点，并且返回链表的头结点。

进阶：你能尝试使用一趟扫描实现吗？


示例 1：
输入：head = [1,2,3,4,5], n = 2
输出：[1,2,3,5]

示例 2：
输入：head = [1], n = 1
输出：[]

示例 3：
输入：head = [1,2], n = 1
输出：[1]


提示：

链表中结点的数目为 sz
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。*/
#include <iostream>

using namespace std;

//Definition for singly-linked list.
struct ListNode {
    int val;
    ListNode *next;

    ListNode() : val(0), next(nullptr) {}

    ListNode(int x) : val(x), next(nullptr) {}

    ListNode(int x, ListNode *next) : val(x), next(next) {}
};

class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        auto *dummy = new ListNode(0, head); // 哑节点 便于删除头节点
        ListNode *fast = head;
        ListNode *slow = dummy;
        for (int i = 0; i < n; ++i) {
            fast = fast->next;
        }
        // 两个指针相差n
        while (fast) {
            fast = fast->next;
            slow = slow->next;
        }
        slow->next = slow->next->next;
        ListNode *ans = dummy->next;
        delete dummy;
        return ans;
    }
};


int main() {
    ListNode n1{1}, n2{2}, n3{3}, n4{4}, n5{5};
    n1.next = &n2;
    n2.next = &n3;
    n3.next = &n4;
    n4.next = &n5;
    Solution sol;
    sol.removeNthFromEnd(&n5, 1);
    return 0;
}